Evaluate: _ x 0 1- 1- - Vidyadip

Question

Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{mx}}{1-\cos\text{nx}}$

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Answer

Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{mx}}{1-\cos\text{nx}}$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{2\sin^{2}\frac{\text{m}}{2}\text{x}}{2\sin^{2}\frac{\text{n}}{2}\text{x}}\bigg)$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\sin\frac{\text{m}}{2}\text{x}}{\sin\frac{\text{n}}{2}\text{x}}\bigg)^{2}$
$=\frac{\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\sin\frac{\text{m}}{2}\text{x}}{\frac{\text{m}}{2}\text{x}}\times\frac{\text{m}}{2}\text{x}\bigg)^{2}}{\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\sin\frac{\text{n}}{2}\text{x}}{\frac{\text{n}}{2}\text{x}}\times\frac{\text{n}}{2}\text{x}\bigg)^{2}}$
$=\frac{1.\frac{\text{m}^{2}}{4}\text{x}^{2}}{1.\frac{\text{n}^{2}}{4}\text{x}^{2}}$
$=\frac{\text{m}^{2}}{\text{n}^{2}}$
Hence, the required answer is $\frac{\text{m}^{2}}{\text{n}^{2}}.$

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